Question

1.Find the capacitance of a parallel plate capacitor having plates of area 5.00 m² that are separated by 0.200 mm of Teflon.

2.What capacitance is needed to store 3.00 µC of charge at a voltage of 120 V?

Answer 1

Here ,

area of plates , A = 5 m^2

seperation of plates , d = 0.2 mm

dielectric constant of teflon , k = 2.1

for a parallel plate capacitor is given as

C = k*epsilon*A/d

C = 2.1 * 8.854 *10^-12 * 5/(.2 *10^-3)

C = 4.65 *10^-7 F

the capacitanc is 4.65 *10^-7 F

2)

as charge = capaitance * V

3 *10^-6 = C * 120

C = 2.5 *10^-8 F

the capacitance is 2.5 *10^-8 F